The figure gives the acceleration of a 5.0 kg particle as an applied force moves
ID: 1773710 • Letter: T
Question
The figure gives the acceleration of a 5.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set by as = 8.0 m/s2. How much work has the force done on the particle when the particle reaches (a) x = 4.0 m, (b) x = 7.0 m, and (c) x = 9.0 m? What is the particle's speed and direction (give positive answer if the particle moves along x axis in positive direction and negative otherwise) of travel when it reaches (d) x = 4.0 m, (e) x = 7.0 m, and (f) x = 9.0 m?
Explanation / Answer
work done = m*area under graph
(a)
work done = m* area of trapezium = 5*(3+4)/2*8 = 140 J
(b)
work done =m*( area of trapezium from x = 0 to x = 5 - area of trapezium from x = 5 to x= 7)
Work done = 5*[ (3+5)*8/2 - (2+1)*8/2) ] = 100 J
(c)
work done =m*( area of trapezium from x = 0 to x = 5 - area of trapezium from x = 5 to x= 9)
Work done = 5*[ (3+5)*8/2 - (4+2)*8/2) ] = 40 J
(d)
from work energy theorem
work done = cahnge in KE
Wa = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)
(1/2)*m*vf^2 = Wa
(1/2)*5*vf^2 = 140
vf = + 7.5 m/s
----------------------
(e)
from work energy theorem
work done = cahnge in KE
Wb = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)
(1/2)*m*vf^2 = Wb
(1/2)*5*vf^2 = 100
vf = + 6.32 m/s
---------------------
(f)
from work energy theorem
work done = cahnge in KE
Wc = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)
(1/2)*m*vf^2 = Wc
(1/2)*5*vf^2 = 40
vf = + 4 m/s